JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Application of Derivatives

From $f(x)=f(1-x)$, the function is symmetric about $x=1/2$.

Option (B): Differentiating $f(x)=f(1-x)$:

At $x=1/2$: $f^{\prime }(1/2)=-f^{\prime }(1/2)$, so $f^{\prime }(1/2)=0$. ✓

Option (A): We have $f^{\prime }(1/4)=0$ and $f^{\prime }(1/2)=0$. Also, from $f^{\prime }(x)=-f^{\prime }(1-x)$, at $x=3/4$: $f^{\prime }(3/4)=-f^{\prime }(1/4)=0$.

By Rolle's theorem, $f^{\prime \prime }(x)=0$ at least once in $(1/4,1/2)$ and at least once in $(1/2,3/4)$. So $f^{\prime \prime }$ vanishes at least twice on $[0,1]$. ✓

Option (C): Let $g(x)=f(x+1/2)$. From $f(x)=f(1-x)$, substituting $x\to x+1/2$:

So $g(x)=g(-x)$, meaning $g$ is even. Therefore $g(x)\sin x$ is odd, and:
✓

Option (D): In the second integral, substitute $u=1-t$:

using $\sin \pi (1-u)=\sin \pi u$. ✓

All four options are correct.